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 chuckfresno
 Joined: 16 Jun 2005  Posts: 39  :   Items 

Posted: Thu Jun 16, 2005 5:44 pm Post subject: Disjoint groups in addition to rows, columns and boxes 


At first blush, the following puzzle seems to have insufficient information to give a unique solution, however, it actually does have one and only one solution and is of medium difficulty:
Code: 
. . .  . . 1  . . .
. 2 .  . 7 .  . 3 .
. . 4  . . .  2 . .
++
8 . .  5 . 6  . . 2
. . .  . 1 .  . . .
1 . .  4 . 7  . . 9
++
. . 3  . . .  1 . .
. 4 .  . 5 .  . 6 .
. . .  9 . .  . . .

Each cell in this Sudoku is a member of FOUR groups rather than the normal THREE (rows, columns and 3x3 boxes). The fourth groups are:
Code: 
a b c  a b c  a b c
d e f  d e f  d e f
g h i  g h i  g h i
++
a b c  a b c  a b c
d e f  d e f  d e f
g h i  g h i  g h i
++
a b c  a b c  a b c
d e f  d e f  d e f
g h i  g h i  g h i

(To be clear, no two cells with the same letter will contain the same digit in the solution.)
This variation is common in several Japanese Number Place magazines. They're printed in color, with each cell in the same disjoint group sharing the same color.
An interesting question is  can the number of clues given for this variant be reduced from the 17 clue record for a standard Sudoku? (18 for a standard symmetric Sudoku). Each clue gives the solver more information, as it interacts with 24 other cells instead of only 20 in a normal Sudoku. 

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 dukuso
 Joined: 14 Jul 2005  Posts: 424  :  Location: germany  Items 

Posted: Thu Jul 14, 2005 8:48 am Post subject: Re: Disjoint groups in addition to rows, columns and boxes 


chuckfresno wrote:  At first blush, the following puzzle seems to have insufficient information to give a unique solution, however, it actually does have one and only one solution and is of medium difficulty:
Code: 
. . .  . . 1  . . .
. 2 .  . 7 .  . 3 .
. . 4  . . .  2 . .
++
8 . .  5 . 6  . . 2
. . .  . 1 .  . . .
1 . .  4 . 7  . . 9
++
. . 3  . . .  1 . .
. 4 .  . 5 .  . 6 .
. . .  9 . .  . . .

Each cell in this Sudoku is a member of FOUR groups rather than the normal THREE (rows, columns and 3x3 boxes). The fourth groups are:
Code: 
a b c  a b c  a b c
d e f  d e f  d e f
g h i  g h i  g h i
++
a b c  a b c  a b c
d e f  d e f  d e f
g h i  g h i  g h i
++
a b c  a b c  a b c
d e f  d e f  d e f
g h i  g h i  g h i

(To be clear, no two cells with the same letter will contain the same digit in the solution.)
This variation is common in several Japanese Number Place magazines. They're printed in color, with each cell in the same disjoint group sharing the same color.
An interesting question is  can the number of clues given for this variant be reduced from the 17 clue record for a standard Sudoku? (18 for a standard symmetric Sudoku). Each clue gives the solver more information, as it interacts with 24 other cells instead of only 20 in a normal Sudoku. 
probably yes. I haven't tried it yet, maybe I'll try it tonight.
Instead of the 2 constraints with blocks, you can require that
no symbol occurs twice in any of the (n+n1) diagonals.
In that case 11 clues are sufficient for a 12*12 puzzle.
For 5,7,11 there are only the trivial doubly cyclic solutions,
for n=2,3,6,8,9,10 there is no solution at all. 

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 chuckfresno
 Joined: 16 Jun 2005  Posts: 39  :   Items 

Posted: Fri Jul 15, 2005 11:28 pm Post subject: Re: Disjoint groups in addition to rows, columns and boxes 


dukuso wrote: 
probably yes. I haven't tried it yet, maybe I'll try it tonight.
Instead of the 2 constraints with blocks, you can require that
no symbol occurs twice in any of the (n+n1) diagonals.
In that case 11 clues are sufficient for a 12*12 puzzle.
For 5,7,11 there are only the trivial doubly cyclic solutions,
for n=2,3,6,8,9,10 there is no solution at all. 
Are you saying that a 12 x 12 puzzle can be made with only 11 clues if all diagonals going one direction are used? (!)
By (n+n1) diagonals, do you mean that the fourth groups would be:
Code: 
a b c d e f g h i j k l
b c d e f g h i j k l a
c d e f g h i j k l a b
d e f g h i j k l a b c
e f g h i j k l a b c d
f g h i j k l a b c d e
g h i j k l a b c d e f
h i j k l a b c d e f g
i j k l a b c d e f g h
j k l a b c d e f g h i
k l a b c d e f g h i j
l a b c d e f g h i j k



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 dukuso
 Joined: 14 Jul 2005  Posts: 424  :  Location: germany  Items 

Posted: Sat Jul 16, 2005 1:22 am Post subject: 


no, what you gave would be called "pandiagonals".
In that case you could probably also make puzzles with n1
clues and a unique solution. But the smallest interesting size
would be 13*13 or 17*17.
My 12*12 puzzle, when solved, would look like a configuration
of 12 sets of 12 chessqueens , with no 2 queens
in any of the sets attacking each other.
See:
http://magictour.free.fr/sudo12.jpg
actually 14 clues, I started with 11 but will have to add further
clues, since it's probably too difficult else.
see also:
http://www.cs.concordia.ca/~chvatal/queengraphs.html
I forgot to examine how many clues are needed for your puzzle,
maybe later... 

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 chuckfresno
 Joined: 16 Jun 2005  Posts: 39  :   Items 

Posted: Sat Jul 16, 2005 4:04 am Post subject: 


dukuso wrote:  no, what you gave would be called "pandiagonals".
In that case you could probably also make puzzles with n1
clues and a unique solution. But the smallest interesting size
would be 13*13 or 17*17.

1) Why n1 clues?
2) Why would the smallest interesting size be that large? What do you mean by "interesting"? Why not 8x8 or 9x9? 

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 dukuso
 Joined: 14 Jul 2005  Posts: 424  :  Location: germany  Items 

Posted: Sat Jul 16, 2005 12:25 pm Post subject: 


chuckfresno wrote:  dukuso wrote:  no, what you gave would be called "pandiagonals".
In that case you could probably also make puzzles with n1
clues and a unique solution. But the smallest interesting size
would be 13*13 or 17*17.

1) Why n1 clues?
2) Why would the smallest interesting size be that large? What do you mean by "interesting"? Why not 8x8 or 9x9? 
1) n1 is the minimum. n2 is not possible since two symbols
were unused and you could always swap these for at least
two different solutions
2) there are pandiagonal latin squares of order n, iff neither
2 nor 3 divides n. So 8*8 or 9*9 are impossible.
All PLS of order <12 are "linear" : A(x,y)=(c*x+d*y+e) mod n
for constants c,d,e. They are easy to construct and to find
since they expand cyclically in both directions 

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 dukuso
 Joined: 14 Jul 2005  Posts: 424  :  Location: germany  Items 

Posted: Sat Jul 16, 2005 8:36 pm Post subject: 


generating random sudokus and then deleting as many clues as possible
such that there is still a unique solution,
I remain with 24 clues in average for the normal sudoku.
For your version with the additional constraint it's 18 clues,
minimum was 16 with 100 tries. 

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 chuckfresno
 Joined: 16 Jun 2005  Posts: 39  :   Items 

Posted: Sat Jul 16, 2005 11:00 pm Post subject: 


I’m confused. Tell me where I misunderstand.
A standard Sudoku solution has 3 groups: rows, columns and boxes.
A common variant published in Japanese’s Number Place magazines includes a fourth group. Call it “disjoint box”:
This type of puzzle has four simultaneous groups:
Code: 
Row Column Box Disjoint box
a a a a a b c d a a b b a b a b
b b b b a b c d a a b b d e d e
c c c c a b c d c c d d a b a b
d d d d a b c d c c d d d e d e

The cells of the disjoint box groups are color coded for easier solving.
This order 4 square has all four groups:
Code: 
1 2 3 4
3 4 1 2
2 1 4 3
4 3 2 1

Here’s an order 9 example:
Code: 
1 2 3 4 5 6 7 8 9
4 5 6 7 8 9 1 2 3
7 8 9 1 2 3 4 5 6
2 3 4 5 6 7 8 9 1
5 6 7 8 9 1 2 3 4
8 9 1 2 3 4 5 6 7
3 4 5 6 7 8 9 1 2
6 7 8 9 1 2 3 4 5
9 1 2 3 4 5 6 7 8

If I understand the way you are using “pandiagonal”, then a pandiagonal square of order five would have these four groups:
Code: 
Rows  Columns  Right and left broken diagonals 
a a a a a a b c d e a b c d e e a b c d
b b b b b a b c d e b c d e a d e a b c
c c c c c a b c d e c d e a b c d e a b
d d d d d a b c d e d e a b c b c d e c
e e e e e a b c d e e a b c d a b c d e

This order 5 square has all four groups:
Code: 
2 3 4 5 1
4 5 1 2 3
1 2 3 4 5
3 4 5 1 2
5 1 2 3 4

Is this different from the 5 sets of 5 nonattacking queens?
Order 9 pandiagonal squares are impossible, right?
Order 11 IS possible:
Order 11 4 group pandiagonal solution:
Code: 
1 2 3 4 5 6 7 8 9 t e
3 4 5 6 7 8 9 t e 1 2
5 6 7 8 9 t e 1 2 3 4
7 8 9 t e 1 2 3 4 5 6
9 t e 1 2 3 4 5 6 7 8
e 1 2 3 4 5 6 7 8 9 t
2 3 4 5 6 7 8 9 t e 1
4 5 6 7 8 9 t e 1 2 3
6 7 8 9 t e 1 2 3 4 5
8 9 t e 1 2 3 4 5 6 7
t e 1 2 3 4 5 6 7 8 9

Ignoring the difficulty that the solver might have keeping track of the broken diagonals, is there another reason that this size could not be used? Is it that there are too few pandiagonal order 11 squares? If it *can* work, it seems like the best size as it's only about 50% larger than the standard size.
dukuso wrote: 
...Instead of the 2 constraints with blocks, you can require that
no symbol occurs twice in any of the (n+n1) diagonals.

What do you mean by “(n+n1) diagonals”.
How is this different from what you call "pandiagonal"?
Are you suggesting that the fourth group be something like:
Code: 
b c d e a
d e a b c
a b c d e
c d e a b
e a b c d

This would seem very difficult for the solver to keep track of, even if the grid was color coded. And if it weren't color coded, is it up to the solver to discover which cell belongs to which disjointnonattackingqueen group?
Digressing, a fifth group can be added to the above order 5 square:
Code: 
+++
 2 3  4 5 1 
 ++ +
 4 5  1  2 3 
 ++ ++
 1  2 3 4  5 
++ ++ 
 3 4  5  1 2 
+ ++ 
 5 1 2  3 4 
+++

An order 7 or 11 puzzle could have polyomino or disjoint groups as well. Polyomino groups are also popular in the Japanese magazines, either instead of the standard box group, or an incomplete set of polyomino and/or disjoint groups in addition to all the normal groups.
For example, this puzzle:
Code: 
. . . . . . 6 . .
. . . . 9 . . . .
. . . 1 4 . . . 7
9 5 . . . . . 4 3
. 6 . . . . . 5 .
4 3 . . . . . 6 8
2 . . . 8 9 . . .
. . . . 2 . . . .
. . 7 . . . . . .

... has the standard row, column and box groups, but cannot be solved without using these four additional groups, each of which must have the digits 19 in the solution:
Code: 
. . a a . . . . .
. a a . . b b b .
a . . . . b b b .
. a a . . b b b .
. . a a . c c . .
. d d d . . c c .
. d d d . . . . c
. d d d . . c c .
. . . . . c c . .



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 chuckfresno
 Joined: 16 Jun 2005  Posts: 39  :   Items 

Posted: Sun Jul 17, 2005 12:48 am Post subject: 


dukuso wrote:  generating random sudokus and then deleting as many clues as possible
such that there is still a unique solution,
I remain with 24 clues in average for the normal sudoku.
For your version with the additional constraint it's 18 clues,
minimum was 16 with 100 tries. 
On average, how often do you get a 17 while generating normal sudoku?
Last edited by chuckfresno on Mon Jul 18, 2005 4:49 am; edited 1 time in total 

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 dukuso
 Joined: 14 Jul 2005  Posts: 424  :  Location: germany  Items 

Posted: Sun Jul 17, 2005 10:22 am Post subject: 


with that program, which is not very fast, I got:
9,19,29,31,9,2,0,0,1 special sudokus with 16,..,24 clues
and 3,30,37,19,9,2 normal sudokus with 22,..,27 clues 

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 Michael Kennett
 Joined: 21 Jul 2005  Posts: 11  :  Location: Adelaide, South Australia  Items 

Posted: Thu Jul 21, 2005 7:51 am Post subject: Re: Disjoint groups in addition to rows, columns and boxes 


[quote="chuckfresno"]At first blush, the following puzzle seems to have insufficient information to give a unique solution, however, it actually does have one and only one solution and is of medium difficulty:
Code: 
. . .  . . 1  . . .
. 2 .  . 7 .  . 3 .
. . 4  . . .  2 . .
++
8 . .  5 . 6  . . 2
. . .  . 1 .  . . .
1 . .  4 . 7  . . 9
++
. . 3  . . .  1 . .
. 4 .  . 5 .  . 6 .
. . .  9 . .  . . .

[/quote="chuckfresno"]
I've just run your board through my solver, and found more than one solution (indeed, got 488!). The first 6 solutions that were generated (with fixed digits in parenthesis) are reproduced below.
Is the original board missing a digit?
Code:  3 8 5  6 2(1) 7 9 4
6(2)9  8(7)4  5(3)1
7 1(4) 3 9 5 (2)8 6
++
(8)9 7 (5)3(6) 4 1(2)
4 5 6  2(1)9  8 7 3
(1)3 2 (4)8(7) 6 5(9)
++
9 6(3) 7 4 8 (1)2 5
2(4)8  1(5)3  9(6)7
5 7 1 (9)6 2  3 4 8
3 8 9  6 2(1) 7 4 5
5(2)6  8(7)4  9(3)1
7 1(4) 3 9 5 (2)8 6
++
(8)9 7 (5)3(6) 4 1(2)
4 6 5  2(1)9  8 7 3
(1)3 2 (4)8(7) 6 5(9)
++
2 5(3) 7 6 8 (1)9 4
9(4)8  1(5)2  3(6)7
6 7 1 (9)4 3  5 2 8
7 8 9  3 2(1) 5 4 6
5(2)6  8(7)4  9(3)1
3 1(4) 6 9 5 (2)8 7
++
(8)9 7 (5)3(6) 4 1(2)
4 6 5  2(1)9  3 7 8
(1)3 2 (4)8(7) 6 5(9)
++
2 5(3) 7 6 8 (1)9 4
9(4)8  1(5)2  7(6)3
6 7 1 (9)4 3  8 2 5
3 8 9  6 2(1) 5 4 7
5(2)6  8(7)4  9(3)1
7 1(4) 3 9 5 (2)8 6
++
(8)9 7 (5)3(6) 4 1(2)
4 6 5  2(1)9  3 7 8
(1)3 2 (4)8(7) 6 5(9)
++
2 5(3) 7 6 8 (1)9 4
9(4)8  1(5)2  7(6)3
6 7 1 (9)4 3  8 2 5
3 8 5  6 2(1) 7 9 4
6(2)9  8(7)4  5(3)1
7 1(4) 3 9 5 (2)8 6
++
(8)9 7 (5)3(6) 4 1(2)
4 5 6  2(1)9  3 7 8
(1)3 2 (4)8(7) 6 5(9)
++
9 6(3) 7 4 8 (1)2 5
2(4)8  1(5)3  9(6)7
5 7 1 (9)6 2  8 4 3
3 8 9  6 2(1) 5 4 7
6(2)5  8(7)4  9(3)1
7 1(4) 3 9 5 (2)8 6
++
(8)9 7 (5)3(6) 4 1(2)
4 5 6  2(1)9  3 7 8
(1)3 2 (4)8(7) 6 5(9)
++
2 6(3) 7 4 8 (1)9 5
9(4)8  1(5)2  7(6)3
5 7 1 (9)6 3  8 2 4 


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 jaap
 Joined: 13 Jun 2005  Posts: 24  :  Location: NL  Items 

Posted: Thu Jul 21, 2005 9:04 am Post subject: Re: Disjoint groups in addition to rows, columns and boxes 


Michael Kennett wrote:  chuckfresno wrote:  At first blush, the following puzzle seems to have insufficient information to give a unique solution, however, it actually does have one and only one solution and is of medium difficulty:

I've just run your board through my solver, and found more than one solution (indeed, got 488!). The first 6 solutions that were generated (with fixed digits in parenthesis) are reproduced below.
Is the original board missing a digit?

Did you read the part in that post about the extra groups? In the first solution you give, there are 2s at (6,3) and (9,6), violating those extra constraints. In the second, (1,7) and (7,4) are both 7s. The other solutions you quote similarly do not satisfy the extra group constraints. _________________ Jaap

Jaap's Puzzle Page:
http://www.geocities.com/jaapsch/puzzles 

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 Michael Kennett
 Joined: 21 Jul 2005  Posts: 11  :  Location: Adelaide, South Australia  Items 

Posted: Thu Jul 21, 2005 10:19 am Post subject: 


Thanks for the clarification  I misunderstood the original posting. 

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 chuckfresno
 Joined: 16 Jun 2005  Posts: 39  :   Items 

Posted: Sat Jul 23, 2005 5:49 am Post subject: 


Why not modify your solver/gererator to handle this type of Sudoku variant? I think you'd be the first.
I really like your idea of using templates to design the look of the puzzles exactly, rather than just symmetrical, but randomly placed, clues that most everyone else uses. This makes it simple to create nonsymmetrical patterns in bulk like:
Code: 
1 4 2 . . . . . .
3 . 7 . . . 5 9 1
6 5 9 . . . 4 . 2
. . . . . . 9 3 6
. 3 1 7 . . . . .
. 9 . 5 . . . . .
. 2 5 8 . 1 6 7 .
. . . . . 3 . 2 .
. . . . . 6 8 5 .

... or one with someone's initials:
Code: 
. 8 . . 1 . . 9 .
. . . . . . . . .
6 . . . 5 . 8 . 3
4 3 . 8 7 . 2 . 6
1 . 6 . 2 . 5 4 .
5 . . . 4 . 3 . 9
2 . . . 8 . 9 . 7
. . . . . . . . .
. 4 . . 3 . . 5 .

Templates might also be used to used to create an entirerly different Sudoku variant:
Code: 
++++
 7  . 3 . 6 . . .  . 
  +++ 
 .  9  .  . . 4 . . 2 
    ++++
 8  .  .  .  .  .  . . . 
 ++    ++ 
 .  . .  6  .  . . 2  . 
  +++ ++  
 9  .  . 2 . 8 .  .  3 
  ++ +++  
 .  2 . .  .  5  . .  . 
 ++    ++ 
 . . .  .  .  .  .  .  4 
++++    
 1 . . 4 . .  .  3  . 
 +++  
 .  . . . 7 . 4 .  6 
++++

Instead of 3x3 boxes, 9 irregularly shaped areas are used. These are a very common variant in some of the Japanese puzzle mgazines.
Both these variations are currently hard to find outside of Japanese publications. 

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 dukuso
 Joined: 14 Jul 2005  Posts: 424  :  Location: germany  Items 

Posted: Sat Jul 23, 2005 7:41 am Post subject: 


>Posted: Sat Jul 16, 2005 11:00 pm Post subject:
>
sorry, I had missed this post.
>
>
>I'm confused. Tell me where I misunderstand.
>
>A standard Sudoku solution has 3 groups: rows, columns and boxes.
>
>A common variant published in Japaneseâ€™s Number Place magazines
>includes a fourth group. Call it â€śdisjoint boxâ€ť:
>
>This type of puzzle has four simultaneous groups:
>Code:
>
>Row Column Box Disjoint box
>a a a a a b c d a a b b a b a b
>b b b b a b c d a a b b d e d e
>c c c c a b c d c c d d a b a b
>d d d d a b c d c c d d d e d e
>
>
>
>The cells of the disjoint box groups are color coded for easier solving.
>
>This order 4 square has all four groups:
>Code:
>
>1 2 3 4
>3 4 1 2
>2 1 4 3
>4 3 2 1
>
>
>
>Hereâ€™s an order 9 example:
>Code:
>
>1 2 3 4 5 6 7 8 9
>4 5 6 7 8 9 1 2 3
>7 8 9 1 2 3 4 5 6
>2 3 4 5 6 7 8 9 1
>5 6 7 8 9 1 2 3 4
>8 9 1 2 3 4 5 6 7
>3 4 5 6 7 8 9 1 2
>6 7 8 9 1 2 3 4 5
>9 1 2 3 4 5 6 7 8
>
>
>
yes.
>
>
>
>If I understand the way you are using â€śpandiagonalâ€ť,
>then a pandiagonal square of order five would have these four groups:
>
>Code:
>
>Rows  Columns  Right and left broken diagonals 
>a a a a a a b c d e a b c d e e a b c d
>b b b b b a b c d e b c d e a d e a b c
>c c c c c a b c d e c d e a b c d e a b
>d d d d d a b c d e d e a b c b c d e c
>e e e e e a b c d e e a b c d a b c d e
>
>
>
>
>This order 5 square has all four groups:
>Code:
>
>2 3 4 5 1
>4 5 1 2 3
>1 2 3 4 5
>3 4 5 1 2
>5 1 2 3 4
>
>
>
>
>Is this different from the 5 sets of 5 nonattacking queens?
it's a collection of 5 disjoint sets of nonattacking (pan) queens
>Order 9 pandiagonal squares are impossible, right?
yes.
>Order 11 IS possible:
yes
>Order 11 4 group pandiagonal solution:
>Code:
>
>1 2 3 4 5 6 7 8 9 t e
>3 4 5 6 7 8 9 t e 1 2
>5 6 7 8 9 t e 1 2 3 4
>7 8 9 t e 1 2 3 4 5 6
>9 t e 1 2 3 4 5 6 7 8
>e 1 2 3 4 5 6 7 8 9 t
>2 3 4 5 6 7 8 9 t e 1
>4 5 6 7 8 9 t e 1 2 3
>6 7 8 9 t e 1 2 3 4 5
>8 9 t e 1 2 3 4 5 6 7
>t e 1 2 3 4 5 6 7 8 9
>
yes
>
>Ignoring the difficulty that the solver might have keeping track
>of the broken diagonals,
no problem for the computer
>is there another reason that this size
>could not be used? Is it that there are too few pandiagonal
>order 11 squares? If it *can* work, it seems like the best
>size as it's only about 50% larger than the standard size.
the reason is, that all rows or columns are just a shifted
version of the 1st row or column. And always bey the same amount.
So this is pretty easy to find.
Every solution of size 11 is of this type.
>dukuso wrote:
>
>
>...Instead of the 2 constraints with blocks, you can require that
>no symbol occurs twice in any of the (n+n1) diagonals.
>
>
>
>What do you mean by â€ś(n+n1) diagonalsâ€ť.
>
>How is this different from what you call "pandiagonal"?
the diagonals don't wrap around.
e f g h i
d e f g h
c d e f g
b c d e f
a b c d e
no two cells with same letter may contain the same digit.
For same diagonals it's only a small constraint, since
it only contains few cells. Only the two main diagonals
contain n cells.
>Are you suggesting that the fourth group be something like:
>
>Code:
>
>b c d e a
>d e a b c
>a b c d e
>c d e a b
>e a b c d
>
no, but that's also interesting ! See Henry D. Shapiro: Generalized
Latin Squares on the Torus. Discrete Math.24 (1978)no.1,6377
>
>This would seem very difficult for the solver to keep track of,
>even if the grid was color coded. And if it weren't color coded,
>is it up to the solver to discover which cell belongs to which
>disjointnonattackingqueen group?
>
>Digressing, a fifth group can be added to the above order 5 square:
>Code:
>
>+++
> 2 3  4 5 1 
> ++ +
> 4 5  1  2 3 
> ++ ++
> 1  2 3 4  5 
>++ ++ 
> 3 4  5  1 2 
>+ ++ 
> 5 1 2  3 4 
>+++
>
yes, but it's not necessary that each set has n elements and that
there are n sets in the group as long as you just only require
that any two digits in a set from a group are different
>
>An order 7 or 11 puzzle could have polyomino or disjoint groups as well.
>Polyomino groups are also popular in the Japanese magazines, either
>instead of the standard box group, or an incomplete set of polyomino
>and/or disjoint groups in addition to all the normal groups.
>
>For example, this puzzle:
>
>Code:
>
>. . . . . . 6 . .
>. . . . 9 . . . .
>. . . 1 4 . . . 7
>9 5 . . . . . 4 3
>. 6 . . . . . 5 .
>4 3 . . . . . 6 8
>2 . . . 8 9 . . .
>. . . . 2 . . . .
>. . 7 . . . . . .
>
>
>
>... has the standard row, column and box groups, but cannot be solved
>without using these four additional groups, each of which must have
>the digits 19 in the solution:
>
>Code:
>
>. . a a . . . . .
>. a a . . b b b .
>a . . . . b b b .
>. a a . . b b b .
>. . a a . c c . .
>. d d d . . c c .
>. d d d . . . . c
>. d d d . . c c .
>. . . . . c c . . o
Guenter. sterten@aol.com 

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